QJEEM23Ma458c: JEE Main 2023 Mathematics - Class 11 JEEM23Ma458c Solution: Let . If and , then is equal to :

We are given that $\left(a+bx+cx^2\right)^{10} = \sum_{i=0}^{20} p_i x^i$ , and we are given that $p_1 = 20$ and $p_2 = 210$ .

We need to find the value of $2(a+b+c)$ .

Using the multinomial theorem, we can express the expansion of $(a + bx + cx^2)^{10}$ as follows:

$\sum\limits_{k_1+k_2+k_3=10} {{10!} \over {{k_1}!{k_2}!{k_3}!}} a^{k_1} (bx)^{k_2} (cx^2)^{k_3}$

Now we need to find the coefficients of $x^1$ and $x^2$ in the expansion:

For $x^1$ term, we have:

$k_2 = 1, k_1 = 9, k_3 = 0$

So,

$p_1 = {{10!} \over {9!1!0!}} a^9 b^1 = 10a^9 b$

For $x^2$ term, there are two possibilities:

$k_2 = 2, k_1 = 8, k_3 = 0 \quad \text{and} \quad k_2 = 0, k_1 = 9, k_3 = 1$

So,

$p_2 = {{10!} \over {8!2!0!}} a^8 b^2 + {{10!} \over {9!0!1!}} a^9 c = 45a^8 b^2 + 10a^9 c$

Now we are given $p_1 = 20$ and $p_2 = 210$ . So, $10a^9 b = 20 \implies a^9 b = 2$

and $45a^8 b^2 + 10a^9 c = 210$

Now, divide the second equation by $a^8$ : $45b^2 + 10ac = 210$

We know that $a^9 b = 2$ . Taking the $9^{th}$ root of both sides: $ab = \sqrt[9]{2}$

Now, let $k = ab = \sqrt[9]{2}$ . We can rewrite the equation for $x^2$ term as: $45k^2 + 10k^9 = 210$

From the equation $ab = k = \sqrt[9]{2}$ , we know that $a$ and $b$ are positive integers. Thus, $k = 2$ (as both $a$ and $b$ must be factors of 2). Now we have:

$a+b = 2$

and from the equation $a^9 b = 2$ , we get $a = 1, b = 2$ or vice versa.

Now we need to find the value of $c$ . We can use the equation for the $x^2$ term again:

$45a^8 b^2 + 10a^9 c = 210$

Using $a=1$ and $b=2$ , we get:

$45(1)^8 (2)^2 + 10(1)^9 c = 210 \implies 180 + 10c = 210 \implies c = 3$

So, $a=1$ , $b=2$ , and $c=3$ . Now, we need to find the value of $2(a+b+c)$ :

$2(a+b+c) = 2(1+2+3) = 2(6) = 12$

Therefore, the answer is $\boxed{12}$ .